Maximum Flow

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Maximum Flow

• Maximum Flow Problem
• The Ford-Fulkerson method
• Maximum bipartite matching

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• material coursing through a system from a source
  to a sink

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Flow networks

• A flow network G(V,E) a directed graph, where
  each edge (u,v)?E has a nonnegative capacity
  c(u,v)gt0.
• If (u,v)?E, we assume that c(u,v)0.
• two distinct vertices a source s and a sink t.

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Flow

• G(V,E) a flow network with capacity function
  c.
• s-- the source and t-- the sink.
• A flow in G a real-valued function fVV ? R
  satisfying the following three properties
• Capacity constraint For all u,v ?V,
• we require f(u,v) ? c( u,v).
• Flow conservation For all u ?V-s,t, we require
  

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Net flow and value of a flow f

• The quantity f (u,v), which can be positive or
  negative, is called the net flow from vertex u to
  vertex v.
• The value of a flow is defined as
• The total flow from source to any other vertices.
• The same as the total flow from any vertices to
  the sink.

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A flow f in G with value .
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Maximum-flow problem

• Given a flow network G with source s and sink t
• Find a flow of maximum value from s to t.
• How to solve it efficiently?

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The Ford-Fulkerson method

• This section presents the Ford-Fulkerson method
  for solving the maximum-flow problem.We call it a
  method rather than an algorithm because it
  encompasses several implementations with
  different running times.The Ford-Fulkerson method
  depends on three important ideas that transcend
  the method and are relevant to many flow
  algorithms and problems residual
  networks,augmenting paths,and cuts. These ideas
  are essential to the important max-flow min-cut
  theorem,which characterizes the value of maximum
  flow in terms of cuts of the flow network.

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Continue

• FORD-FULKERSON-METHOD(G,s,t)
• initialize flow f to 0
• while there exists an augmenting path p
• do augment flow f along p
• return f

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Residual networks

• Given a flow network and a flow, the residual
  network consists of edges that can admit more net
  flow.
• G(V,E) --a flow network with source s and sink
  t
• f a flow in G.
• The amount of additional net flow from u to v
  before exceeding the capacity c(u,v) is the
  residual capacity of (u,v), given by
  cf(u,v)c(u,v)-f(u,v)

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Example of residual network
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s
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(a)
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Example of Residual network (continued)
s
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(b)
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Fact 1

• Let G(V,E) be a flow network with source s and
  sink t, and let f be a flow in G.
• Let Gf be the residual network of G induced by
  f,and let f be a flow in Gf.Then, the flow sum
  ff is a flow in G with value
  .

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Augmenting paths

• Given a flow network G(V,E) and a flow f, an
  augmenting path is a simple path from s to t in
  the residual network Gf.
• Residual capacity of p the maximum amount of
  net flow that we can ship along the edges of an
  augmenting path p, i.e., cf(p)mincf(u,v)(u,v)
  is on p.

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The residual capacity is 1.
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Example of an augment path (bold edges)
s
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(b)
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The basic Ford-Fulkerson algorithm

• FORD-FULKERSON(G,s,t)
• for each edge (u,v) ?EG
• do fu,v 0
• fv,u 0
• while there exists a path p from s to t in the
  residual network Gf
• do cf(p) mincf(u,v) (u,v) is in
  p
• for each edge (u,v) in p
• do fu,v fu,vcf(p)

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Example

• The execution of the basic Ford-Fulkerson
  algorithm.(a)-(d) Successive iterations of the
  while loop. The left side of each part shows the
  residual network Gf from line 4 with a shaded
  augmenting path p.The right side of each part
  shows the new flow f that results from adding fp
  to f.The residual network in (a) is the input
  network G.(e) The residual network at the last
  while loop test.It has no augmenting paths,and
  the flow f shown in (d) is therefore a maximum
  flow.

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(a)
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s
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(b)
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(c)
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(d)
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(e)
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Time complexity

• If each c(e) is an integer, then time complexity
  is O(Ef), where f is the maximum flow.
• Reason each time the flow is increased by at
  least one.
• This might not be a polynomial time algorithm
  since f can be represented by log (f) bits. So,
  the input size might be log(f).

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Cuts of flow networks

• The proof of the correctness of the
  Ford-Fulkerson method depends on a concept cut.
  
• A cut (S,T) of flow network G(V,E) is a
  partition of V into S and TV-S such that s?S and
  t ?T.
• If f is a flow, then the net flow across the cut
  (S,T) is F(S,T)? u?Sv?T f(u, v).
• The capacity of the cut (S,T) is
• c(S, T) ? u?Sv?T c(u, v).

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S
T
A cut (S,T), where Ss,v1,v2 and Tv3,v4,t.
The net flow across (S,T) is f(S,T)12-41119,and
the capacity is c(S,T)121426.
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Continue

• Lemma 27.5
• Let f be a flow in a flow network G with source s
  and sink t,and let (S,T) be a cut of G.Then, the
  net flow across (S,T) is f(S,T) .
• Proof 1. f(S-s, V)0 by flow conservation.
• 2. f(S, S)0 since f(u, v)-f(v, u).
• f(S, T)f(S, V)-f(S, S)f(S, V)
• f(s, V)f(S-s, V)f(s, V)f.

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Continue

• Corollary 27.6
• The value of any flow f in a flow network G is
  bounded from above by the capacity of any cut of
  G.
• Proof f(S, T)?c(S, T).

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• Theorem 27.7(Max-flow min-cut theorem)
• If f is a flow in a flow network G(V,E) with
  source s and sink t,then the following conditions
  are equivalent
• f is a maximum flow in G
• The residual network Gf contains no augmenting
  paths
• f c(S,T) for some cut (S,T) of G.
• Proof 1?2 Otherwise, if a aug. path exists, we
  can further increase the flow.
• 2?3. If no aug. path exists, then we
  construct S as the set of vertices that is
  reachable from s. TV-S. By construction, there
  is no edge (u, v) in the residual graph such that
  u?S and v?T. Thus, ff(S,T)c(S, T).
• 3?1 ff(S, T)c(S,T). By 27.6,f f(S,
  T)?c(S,T). Thus, f is maximum.

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The Edmonds-Karp algorithm

• Find the augmenting path using breadth-first
  search.
• Breadth-first search gives the shortest path for
  graphs (Assuming the length of each edge is 1.)
• Time complexity of Edmonds-Karp algorithm is
  O(VE2).
• The proof is not required.

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Maximum bipartite matching

• Bipartite graph a graph (V, E), where VL?R,
  LnRempty, and for every (u, v)?E, u ?L and v ?R.
• Given an undirected graph G(V,E), a matching is
  a subset of edges M?E such that for all vertices
  v?V,at most one edge of M is incident on v.We say
  that a vertex v ?V is matched by matching M if
  some edge in M is incident on votherwise, v is
  unmatched. A maximum matching is a matching of
  maximum cardinality,that is, a matching M such
  that for any matching M, we have
  .

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R
(a)
A bipartite graph G(V,E) with vertex partition
VL?R.(a)A matching with cardinality 2.(b) A
maximum matching with cardinality 3.
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Finding a maximum bipartite matching

• We define the corresponding flow network
  G(V,E) for the bipartite graph G as follows.
  Let the source s and sink t be new vertices not
  in V, and let VV?s,t.If the vertex partition
  of G is VL ?R, the directed edges of G are
  given by E(s,u)u?L ?(u,v)u ?L,v ?R,and
  (u,v) ?E ?(v,t)v ?R.Finally, we assign unit
  capacity to each edge in E.
• We will show that a matching in G corresponds
  directly to a flow in Gs corresponding flow
  network G. We say that a flow f on a flow
  network G(V,E) is integer-valued if f(u,v) is an
  integer for all (u,v) ?VV.

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s
t
L
R
(b)
(a)The bipartite graph G(V,E) with vertex
partition VL?R. A maximum matching is shown by
shaded edges.(b) The corresponding flow
network.Each edge has unit capacity.Shaded edges
have a flow of 1,and all other edges carry no
flow.
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Continue

• Lemma 27.10
• Let G(V,E) be a bipartite graph with vertex
  partition VL?R,and let G(V,E) be its
  corresponding flow network.If M is a matching in
  G, then there is an integer-valued flow f in G
  with value .Conversely, if f is
  an integer-valued flow in G,then there is a
  matching M in G with cardinality .
  
• Reason The edges incident to s and t ensures
  this.